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Balancing chemical equations is a fundamental skill in chemistry that ensures the same number of each type of atom appears on both sides of a given reaction. This is crucial for reflecting the conservation of mass during chemical reactions. In the given problem, we have two reactions to deal with, and both must be balanced correctly.

To balance a chemical equation, follow these steps:

• Write down the unbalanced equation.
• Count the number of each type of atom on both sides of the equation.
• Add coefficients in front of chemical formulas to balance the atoms, starting with the most complex molecules.
• Double-check that all atoms balance and that the coefficients are in the simplest ratio.

For example, the first reaction given is: \( \mathrm{2NO + O_{2} \rightarrow 2NO_{2}} \). Here, there are 2 nitrogen atoms and 4 oxygen atoms on both sides, ensuring balance. In the second reaction, \( \mathrm{2NO_{2} + O_{2} \rightarrow 2NO + O_{3}} \), the same principle applies. Balancing these correctly is vital for accurate calculations later on.

Molar mass is crucial for converting between the mass of a substance and the amount of moles it represents. It's calculated by adding the atomic masses of all the atoms within a molecule, according to the periodic table values. This number is expressed in grams per mole (g/mol), providing a bridge between macroscopic and molecular scales.

For the compound nitrogen monoxide (NO), the molar mass calculation is simple:

• Nitrogen (N) has an atomic mass of approximately 14.01 g/mol.
• Oxygen (O) has an atomic mass of approximately 16.00 g/mol.
• Add these together for NO: \( 14.01 + 16.00 = 30.01\, g/mol \)

Knowing this, you can convert 4.55 grams of NO to moles, yielding \( \mathrm{Moles\, of\, NO = \frac{4.55\, g}{30.01\, g/mol} = 0.1516\, mol} \). This is vital for progressing to calculate theoretical yields.

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It tells us how much of each substance is involved. To tackle stoichiometry problems, like the one in this exercise, you need the balanced equation and an understanding of mole ratios.

Start by determining the amount in moles of a known reactant, using its molar mass, as previously explained. Next, apply the coefficients from the balanced equations as mole ratios to find other reactants or products.

In our case, after finding the moles of NO, we use the equation \( \mathrm{2NO + O_{2} \rightarrow 2NO_{2}} \), which tells us that NO and NO2 have a 1:1 ratio. Hence, \( 0.1516\, mol \) of NO produces \( 0.1516\, mol \) of NO2.

For the second reaction, the equation \( \mathrm{2NO_{2} + O_{2} \rightarrow 2NO + O_{3}} \) signifies a molar ratio of 2:1 for NO2 to O3, giving us \( \mathrm{Moles\, of\, O_{3} = \frac{0.1516\, mol\, of\, NO_{2}}{2} = 0.0758\, mol} \). Finally, converting to grams using the molar mass of O3, \(48.00\, g/mol\), calculates the theoretical yield: \( \mathrm{0.0758\, mol \times 48.00\, g/mol = 3.6384\, g} \). This stepwise approach links the parts of stoichiometry to provide a thorough understanding of the chemical reaction process.